Random Resolution Refutations ak and Neil Thapen 1 Pavel Pudl - - PowerPoint PPT Presentation

Random Resolution Refutations

Pavel Pudl´ ak and Neil Thapen1

Mathematical Institute, Academy of Sciences, Prague

Riga, 6.7.17 1the authors are supported by the ERC grant FEALORA [1]

History and Motivation

Stefan Dantchev [unpublished] Buss, Ko lodziejczyk, Thapen [2014]

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History and Motivation

Stefan Dantchev [unpublished] Buss, Ko lodziejczyk, Thapen [2014]

◮ Separations of fragments of Bounded Arithmetic ◮ The first randomized version of a proof system ◮ Developing lower bound methods ◮ Understanding what can be proved from random tautologies

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• Question. Random 3-DNFs (of sufficient density)
• 1. are tautologies,
• 2. can be easily generated, and
• 3. seem to be hard for every proof system (for not too high density).

But can we derive from them any useful tautology?

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• Question. Random 3-DNFs (of sufficient density)
• 1. are tautologies,
• 2. can be easily generated, and
• 3. seem to be hard for every proof system (for not too high density).

But can we derive from them any useful tautology?

Corollary (of our results)

With high probability for a random 3-DNF Φ (of sufficient density), there is no bounded depth Frege proof of Φ → PHP of subexponential size.

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Overview

• 1. equivalent definitions
• 2. upper bounds
• 3. lower bounds
• 4. generalization to bounded depth Frege proofs
• 5. problems

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Definitions

Definition

An ǫ-random resolution distribution, or ǫ-RR distribution, of F is a probability distribution ∆ on pairs (Bi, Πi)i∼∆ such that

• 1. for each i ∈ ∆, Bi is a CNF in variables x1, . . . , xn and Πi is a

resolution refutation of F ∧ Bi

• 2. for every α ∈ {0, 1}n, Pri∼∆[Bi is satisfied by α] ≥ 1 − ǫ.

The size and the width of ∆ are defined respectively as the maximum size and maximum width of the refutations Πi (if these maxima exist).

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◮ RR is sound as a refutational system, in the sense that if F has an

ǫ-RR distribution then F is unsatisfiable. Proof: consider any assignment α ∈ {0, 1}n. Since ǫ < 1, there is at least one pair (Bi, Πi) such that α satisfies Bi and Πi is a resolution refutation of F ∧ Bi. So α cannot also satisfy F, by the soundness

• f resolution.

◮ RR is complete, since resolution is complete. ◮ RR is not a propositional proof system in the sense of Cook and

Reckhow because it is defined by a semantic condition.

◮ The error ǫ can be reduced with only moderate increase of the

proofs, so we can w.l.o.g. assume ǫ = 1/2.

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Definition

Let ∆ be a probability distribution on {0, 1}n. An (ǫ, ∆)-random resolution refutation, or (ǫ, ∆)-RR refutation, of F is a pair (B, Π) such that

• 1. B is a CNF in variables x1, . . . , xn and Π is a resolution refutation of

F ∧ B

• 2. Prα∼∆[B[α] = 1] ≥ 1 − ǫ.

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Definition

Let ∆ be a probability distribution on {0, 1}n. An (ǫ, ∆)-random resolution refutation, or (ǫ, ∆)-RR refutation, of F is a pair (B, Π) such that

• 1. B is a CNF in variables x1, . . . , xn and Π is a resolution refutation of

F ∧ B

• 2. Prα∼∆[B[α] = 1] ≥ 1 − ǫ.

If an (ǫ, ∆)-RR refutation exists for all distributions ∆, then this is equivalent to the existence of an ǫ-RR distribution.

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Let P be a class of resolution refutations, e.g., refutations of width w and size s for some w and s.

Proposition

The following are equivalent.

• 1. F has an ǫ-RR distribution of refutations from P.
• 2. F has an (ǫ, ∆)-RR refutation from P for every distribution ∆ on

{0, 1}n.

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Proof.

Consider a zero-sum game between two players, Prover and Adversary:

◮ Prover picks a pair (B, Π) such that Π ∈ P, B is a CNF, and Π is a

refutation of F ∧ B,

◮ Adversary picks an assignment α.

The payoff is B[α], i.e., Prover gets 1 if α satisfies B and 0 otherwise. Then

◮ definition 1 says: Prover has a mixed strategy to achieve a payoff of

at least 1 − ǫ, and

◮ definition 2 says: Adversary does not have a mixed strategy to

achieve a payoff less than 1 − ǫ. By the minimax theorem these statements are equivalent.

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Definition

Let A ⊆ {0, 1}n be a nonempty set of truth assignments. We say that a formula C is a semantic consequence over A of formulas C1, . . . , Cr, if every assignment in A that satisfies C1, . . . , Cr also satisfies C. A semantic resolution refutation of F over A is a sequence Π of clauses, ending with the empty clause, in which every clause either belongs to F

• r is a semantic consequence over A of at most two earlier clauses.

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Definition

Let A ⊆ {0, 1}n be a nonempty set of truth assignments. We say that a formula C is a semantic consequence over A of formulas C1, . . . , Cr, if every assignment in A that satisfies C1, . . . , Cr also satisfies C. A semantic resolution refutation of F over A is a sequence Π of clauses, ending with the empty clause, in which every clause either belongs to F

• r is a semantic consequence over A of at most two earlier clauses.

Definition

Let ∆ be a probability distribution on {0, 1}n. An (ǫ, ∆)-semantic refutation of F is a pair (A, Π) such that

• 1. Π is a semantic refutation of F over A, and
• 2. Prα∼∆[α ∈ A] ≥ 1 − ǫ.

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Definition

Let A ⊆ {0, 1}n be a nonempty set of truth assignments. We say that a formula C is a semantic consequence over A of formulas C1, . . . , Cr, if every assignment in A that satisfies C1, . . . , Cr also satisfies C. A semantic resolution refutation of F over A is a sequence Π of clauses, ending with the empty clause, in which every clause either belongs to F

• r is a semantic consequence over A of at most two earlier clauses.

Definition

Let ∆ be a probability distribution on {0, 1}n. An (ǫ, ∆)-semantic refutation of F is a pair (A, Π) such that

• 1. Π is a semantic refutation of F over A, and
• 2. Prα∼∆[α ∈ A] ≥ 1 − ǫ.

Note: no auxiliary formulas!

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Proposition

• 1. If F has an (ǫ, ∆)-RR refutation of width w and size s, then it also

has an (ǫ, ∆)-semantic resolution refutation of width ≤ w and size ≤ s.

• 2. If F has has an (ǫ, ∆)-semantic refutation of width w and size s,

then it also has an (ǫ, ∆)-RR refutation of width O(w) and size at most O(sw 2).

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The strength of RR

◮ A random 3-CNF with n variables and 64n clauses has a 1/2-RR

distribution of constant width and constant size with probability exponentially close to 1.

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The strength of RR

◮ A random 3-CNF with n variables and 64n clauses has a 1/2-RR

distribution of constant width and constant size with probability exponentially close to 1.

◮ The retraction weak pigeonhole principle that asserts that there is

no pair of functions f : [2n] → [n] and g : [n] → [2n] such that g(f (x)) = x for all x < n has a narrow 1/2-RR distribution.

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The strength of RR

◮ A random 3-CNF with n variables and 64n clauses has a 1/2-RR

distribution of constant width and constant size with probability exponentially close to 1.

◮ The retraction weak pigeonhole principle that asserts that there is

no pair of functions f : [2n] → [n] and g : [n] → [2n] such that g(f (x)) = x for all x < n has a narrow 1/2-RR distribution.

◮ If P = NP, then 1/2-RR cannot be polynomially simulated by any

Cook-Reckhow refutation system. In particular, 1/2-RR is not itself a Cook-Reckhow refutation system if P = NP.

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Lemma

Let F := C1 ∧ · · · ∧ Cm be a k-CNF formula such that for every assignment α the number of clauses that are satisfied by α is ≤ δm for some constant 0 < δ < 1. Then F has a δ-RR distribution of size 2k which can be constructed in polynomial time.

Proof.

The distribution is defined by:

• 1. pick i ∈ [m] randomly
• 2. let Bi (the auxiliary formula) be ¬Ci and Πi the proof of ⊥ from Bi

and Ci.

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Lemma

Let F := C1 ∧ · · · ∧ Cm be a k-CNF formula such that for every assignment α the number of clauses that are satisfied by α is ≤ δm for some constant 0 < δ < 1. Then F has a δ-RR distribution of size 2k which can be constructed in polynomial time.

Proof.

The distribution is defined by:

• 1. pick i ∈ [m] randomly
• 2. let Bi (the auxiliary formula) be ¬Ci and Πi the proof of ⊥ from Bi

and Ci. For random 3-CNFs, δ = 7/8. To get ǫ ≤ 1/2, take random sixtuples i1, . . . , i6 ∈ [m] and the CNFs equivalent to ¬Ci1 ∨ · · · ∨ ¬Ci6

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The weakness of RR

Theorem

PHPn has no 1/2-RR distribution of size O(2n1/12).

Theorem

The formula CPLS2

n (will be defined later) does not have a 1/2-RR

distribution of size O(2n1/17). CPLS2

n has polynomial size Res(2) proofs.

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Bounded depth Frege refutation systems

• 1. clauses, -formulas – Resolution = 1-Frege system
• 2. DNFs – -formulas – 2-Frege system
• 3. etc.

Problem

Does there exist a CNF contradiction refutable in some d-Frege system, d > 2, by quasipolynomial size refutation that does not have such a 2-Frege refutation.2

2Open even for 1.5-Frege (=Res(log)). [15]

Bounded depth Frege refutation systems

• 1. clauses, -formulas – Resolution = 1-Frege system
• 2. DNFs – -formulas – 2-Frege system
• 3. etc.

Problem

Does there exist a CNF contradiction refutable in some d-Frege system, d > 2, by quasipolynomial size refutation that does not have such a 2-Frege refutation.2 WPHP separates Resolution from 2-Frege.

2Open even for 1.5-Frege (=Res(log)). [15]

Bounded depth Frege refutation systems

• 1. clauses, -formulas – Resolution = 1-Frege system
• 2. DNFs – -formulas – 2-Frege system
• 3. etc.

Problem

Does there exist a CNF contradiction refutable in some d-Frege system, d > 2, by quasipolynomial size refutation that does not have such a 2-Frege refutation.2 WPHP separates Resolution from 2-Frege. Our result: CPLS2 separates RR from 2-Frege.

2Open even for 1.5-Frege (=Res(log)). [15]

Definition

The formula CPLSa,b,c consists of the following three sets of clauses:

• 1. For each y < c, the clause

¬G0(0, y)

• 2. For each i < a − 1, each pair x, x′ < b and each y < c, the clause

fi(x) = x′ ∧ Gi+1(x′, y) → Gi(x, y)

• 3. For each x < b and each y < c, the clause

u(x) = y → Ga−1(x, y).

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Definition

The formula CPLSa,b,c consists of the following three sets of clauses:

• 1. For each y < c, the clause

¬G0(0, y)

• 2. For each i < a − 1, each pair x, x′ < b and each y < c, the clause

fi(x) = x′ ∧ Gi+1(x′, y) → Gi(x, y)

• 3. For each x < b and each y < c, the clause

u(x) = y → Ga−1(x, y). The formula CPLS2 is a variant of CPLS where for each i, x, y, instead

• f the single variable Gi(x, y) it has two variables G 0

i (x, y) and G 1 i (x, y).

To express that colour y is present at node (i, x) we now use the conjunction G 0

i (x, y) ∧ G 1 i (x, y).

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Lower bound for the pigeonhole principle in RR

PHPn has variables pij for i ∈ U and j ∈ V and consists of clauses 1.

j∈V pij for all i ∈ U

• 2. ¬pij ∨ ¬pi′j for all i, i′ ∈ U with i = i′ and all j ∈ V .

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• 1. width reduction

For a clause C, we define wec(C), the edge covering width or ec-width of C, to be the smallest size of a set W ⊆ U ∪ V that intersects all pairs {i, j} mentioned in C. Formally, wec(C) := min{|W | | ∀i ∈ U, j ∈ V , (pij ∈ C∨¬pij ∈ C) → (i ∈ W ∨j ∈ W )}. If Φ is a CNF formula, then wec(Φ) is the maximum of the ec-widths of its clauses. We will denote by Rm the set of partial matchings of size n − m equipped with the uniform distribution.

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• 1. width reduction

For a clause C, we define wec(C), the edge covering width or ec-width of C, to be the smallest size of a set W ⊆ U ∪ V that intersects all pairs {i, j} mentioned in C. Formally, wec(C) := min{|W | | ∀i ∈ U, j ∈ V , (pij ∈ C∨¬pij ∈ C) → (i ∈ W ∨j ∈ W )}. If Φ is a CNF formula, then wec(Φ) is the maximum of the ec-widths of its clauses. We will denote by Rm the set of partial matchings of size n − m equipped with the uniform distribution.

Lemma

There exist constants c > 0 and 0 < d < 1 such that for every clause C and every 1 ≤ ℓ ≤ n1/2, Pr[wec(C ρ) > ℓ] ≤ dℓ, where the probability is over ρ ∼ R⌊cn1/4⌋.

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• 2. the fixing lemma

Ideally, we would like to eliminate the auxiliary formula B by finding ρ ∈ Rm such that Bρ ≡ 1. This is not always possible.

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• 2. the fixing lemma

Ideally, we would like to eliminate the auxiliary formula B by finding ρ ∈ Rm such that Bρ ≡ 1. This is not always possible.

Lemma

Let B be a CNF formula such that wec(B) ≤ ℓ and Prρ∼Rm[Bρ = 0] ≤ 1/2 where ℓ < m < n. Suppose that ℓm(m − 1) n − m + 1 < 1 2. Then there exists a ρ ∈ Rm such that there is no extension σ ⊇ ρ to a partial matching such that Bσ = 0.

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Random Bounded Depth Frege proofs

Definition

Let d ≥ 3 be constant. A random d-Frege proof of a DNF tautology F is a depth d Frege proof of B → F, where B is a depth 3 formula

i Bi, where Bis are CNFs, such that for

every assignment α at least 1/2 of Bis are satisfied.

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Random Bounded Depth Frege proofs

Definition

Let d ≥ 3 be constant. A random d-Frege proof of a DNF tautology F is a depth d Frege proof of B → F, where B is a depth 3 formula

i Bi, where Bis are CNFs, such that for

every assignment α at least 1/2 of Bis are satisfied.

Theorem

For every d, there is ǫd > 0 such that every random d-Frege proof of PHPn has size ≥ 2nǫd .

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Corollary

With high probability for a random 3-DNF Φ with n variables and 64n clauses, there is no bounded depth Frege proof of Φ → PHP of subexponential size.

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Problems

◮ find natural random version for more proof systems ◮ cutting planes, ResLin, and, in particular, for the Nullstellensatz

system

◮ randomized SAT algorithms and random resolution

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THANK YOU

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